The coordinates of the point on the curve sqr | vedclass

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Question

(C) Given the curve equation is $\sqrt{x}+\sqrt{y}=6$. Differentiating both sides with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y

Vedclass · Accepted Answer

$(9,9)$

Vedclass · Answer

(C) Given the curve equation is $\sqrt{x}+\sqrt{y}=6$. Differentiating both sides with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$. This implies $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$. Since the tangent is equally inclined to the axes,its slope must be $\pm 1$. Thus,$-\frac{\sqrt{y}}{\sqrt{x}} = \pm 1$,which gives $\sqrt{y} = \pm \sqrt{x}$. Since $x$ and $y$ must be positive for the square roots to be defined,we have $\sqrt{y} = \sqrt{x}$,which implies $y = x$. Substituting $y = x$ into the original equation: $\sqrt{x} + \sqrt{x} = 6 2\sqrt{x} = 6 \sqrt{x} = 3 x = 9$. Since $y = x$,we get $y = 9$. Therefore,the required point is $(9, 9)$.

The coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=6$ at which the tangent is equally inclined to the axes are

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